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anonymous

  • one year ago

can't understand why integral of xe^((x^2)/2) is e^((x^2)/2) can someone explain how to calculate this integral please?

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  1. UsukiDoll
    • one year ago
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    \[\LARGE xe^{\frac{x^2}{2}}\] this one?

  2. UsukiDoll
    • one year ago
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    do you know u-substitution?

  3. UsukiDoll
    • one year ago
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    The u-substitution is an integration method that essentially involves using the chain rule in reverse.

  4. UsukiDoll
    • one year ago
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    hmm let's try rewriting that fraction \[\LARGE xe^{\frac{1}{2}x^2}\] so we can see the exponent clearly.

  5. UsukiDoll
    • one year ago
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    we need to pick a u so that when we take that derivative of u we can substitute.. so what will be our u? Well we do see an x in front of the e.. Maybe we can have \[\LARGE u=\frac{1}{2}x^2 \]

  6. UsukiDoll
    • one year ago
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    so what is the derivative of \[\large \frac{1}{2}x^2 ? \]

  7. UsukiDoll
    • one year ago
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    we can use product rule to make this a bit easier. Let f(x) = 1/2 and g(x) = x^2 recall that the product rule is f(x)g'(x) +f'(x)g(x)

  8. UsukiDoll
    • one year ago
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    so what is the derivative of f(x) = 1/2 ?

  9. UsukiDoll
    • one year ago
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    similarly what is the derivative of \[\large g(x) = x^2 \]

  10. anonymous
    • one year ago
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    so if \[u= (1/2)x^2 -> u' = x\]

  11. UsukiDoll
    • one year ago
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    yeah but we have to write this differently it will be du = x dx

  12. anonymous
    • one year ago
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    so now we have \[xe^u dx\]

  13. UsukiDoll
    • one year ago
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    hmm how do I explain this.... we just have \[e^u \] and substitute the u back .. don't forget to add +C

  14. UsukiDoll
    • one year ago
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    remember du = x dx ? since there's nothing on the left hand side.... oh man .. I'm having trouble explaining this part. . . usually when there's nothing ... it's just blank like 1 sorry I'm in panic mode I can't figure out how to type this well

  15. UsukiDoll
    • one year ago
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    but we just have \[\LARGE e^u +C \] now substitute \[\LARGE u=\frac{1}{2}x^2 \] back into \[\LARGE e^u +C \]

  16. anonymous
    • one year ago
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    ok, so \[xe^u dx -> xe^u (1/x) dy -> e^u du\] (+c...)

  17. anonymous
    • one year ago
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    du

  18. UsukiDoll
    • one year ago
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    you don't have to write dx you don't have to write du

  19. UsukiDoll
    • one year ago
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    du = x dx it's just .... @ganeshie8 can you help me explain this? I know what's going on just can't put it into words

  20. anonymous
    • one year ago
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    Ok, I think I understand, thank you very much!

  21. UsukiDoll
    • one year ago
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    but in the end you should have \[\LARGE e^u +C \] now plug in u back into this equation

  22. UsukiDoll
    • one year ago
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    and once we have our antiderivative, we can simply check by taking the derivative.

  23. UsukiDoll
    • one year ago
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    you should have something like \[\LARGE e^{\frac{1}{2}x^2} +C\]

  24. UsukiDoll
    • one year ago
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    now take the derivative of that equation...

  25. UsukiDoll
    • one year ago
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    that's how to check :)

  26. anonymous
    • one year ago
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    so we get back to \[e^((x^2)/2) \]

  27. anonymous
    • one year ago
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    yes, +c

  28. UsukiDoll
    • one year ago
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    mhm

  29. UsukiDoll
    • one year ago
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    if you take the derivative of your antiderivative, we should be getting that x in front of that e

  30. anonymous
    • one year ago
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    Yes, I wrote the integral

  31. UsukiDoll
    • one year ago
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    \[\LARGE f(x) = e^{\frac{1}{2}x^2} +C \] \[\LARGE f'(x) = xe^{\frac{1}{2}x^2} +C \] because the derivative is \[\frac{1}{2}x^2 \rightarrow \frac{1}{2} (2x) + 0(x^2) \rightarrow x\] and that's placed in front of the e.

  32. UsukiDoll
    • one year ago
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    the derivative of any e^x is always in the front and the exponent part of e^x just stays there :)

  33. anonymous
    • one year ago
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    Excellent. Thanks a lot!

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